Solve 18x^2-25x-6

$$18 x ^ { 2 } - 25 x - 6$$

Factor

$$18\left(x-\frac{25-\sqrt{1057}}{36}\right)\left(x-\frac{\sqrt{1057}+25}{36}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$18x^{2}-25x-6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 18\left(-6\right)}}{2\times 18}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 18\left(-6\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-25\right)±\sqrt{625-72\left(-6\right)}}{2\times 18}$$

Multiply -72 times -6.

$$x=\frac{-\left(-25\right)±\sqrt{625+432}}{2\times 18}$$

Add 625 to 432.

$$x=\frac{-\left(-25\right)±\sqrt{1057}}{2\times 18}$$

The opposite of -25 is 25.

$$x=\frac{25±\sqrt{1057}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{25±\sqrt{1057}}{36}$$

Now solve the equation x=\frac{25±\sqrt{1057}}{36} when ± is plus. Add 25 to \sqrt{1057}.

$$x=\frac{\sqrt{1057}+25}{36}$$

Now solve the equation x=\frac{25±\sqrt{1057}}{36} when ± is minus. Subtract \sqrt{1057} from 25.

$$x=\frac{25-\sqrt{1057}}{36}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{25+\sqrt{1057}}{36} for x_{1} and \frac{25-\sqrt{1057}}{36} for x_{2}.

$$18x^{2}-25x-6=18\left(x-\frac{\sqrt{1057}+25}{36}\right)\left(x-\frac{25-\sqrt{1057}}{36}\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 18

$$x ^ 2 -\frac{25}{18}x -\frac{1}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{25}{18} rs = -\frac{1}{3}$$

Two numbers r and s sum up to \frac{25}{18} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{18} = \frac{25}{36}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{25}{36} - u s = \frac{25}{36} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}

$$(\frac{25}{36} - u) (\frac{25}{36} + u) = -\frac{1}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{625}{1296} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{625}{1296} on both sides

$$-u^2 = -\frac{1}{3}-\frac{625}{1296} = -\frac{1057}{1296}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1057}{1296} u = \pm\sqrt{\frac{1057}{1296}} = \pm \frac{\sqrt{1057}}{36}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{25}{36} - \frac{\sqrt{1057}}{36} = -0.209 s = \frac{25}{36} + \frac{\sqrt{1057}}{36} = 1.598$$

Evaluate

$$18x^{2}-25x-6$$